SOLVING TRIGONOMETRIC EQUATIONS


Note:

If you would like an review of trigonometry, click on trigonometry.


Solve for x in the following equation.


Example 2:        

$7\sin \left( 5x\right) -1=0$


There are an infinite number of solutions to this problem. To solve for x, you must first isolate the sine term.

\begin{eqnarray*}&& \\ 7\sin \left( 5x\right) -1 &=&0 \\ && \\ \sin \left( 5x\right) &=&\displaystyle \frac{1}{7} \\ && \\ && \end{eqnarray*}

If we restriction the domain of the sine function to $\left[ -\displaystyle \frac{\pi }{2} \leq 5x\leq ,\displaystyle \frac{\pi }{2}\... ...-\displaystyle \frac{\pi }{10}\leq x\leq ,\displaystyle \frac{\pi }{10}\right] $, we can use the inverse sine function to solve for reference angle 5x and then x.

\begin{eqnarray*}&& \\ \sin \left( 5x\right) &=&\displaystyle \frac{1}{7} \\ &... ...ht) &=&\sin ^{-1}\left( \displaystyle \frac{1}{7 }\right) \\ && \end{eqnarray*}
\begin{eqnarray*}&&\\ 5x &=&\sin ^{-1}\left( \displaystyle \frac{1}{7}\right) \... ...x &\approx &0.028669513781\ \mbox{ radians }\\ && \\ && \\ && \end{eqnarray*}

We know that the $\sin $e function is positive in the first and the second quadrant. Therefore two of the solutions are the angle 5x that terminates in the first quadrant and the angle $\pi -5x$ that terminates in the second quadrant. We have already solved for 5x.

\begin{eqnarray*}&& \\ \sin \left( \pi -5x\right) &=&\displaystyle \frac{1}{7} ... ... -5x &=&\sin ^{-1}\left( \displaystyle \frac{1}{7}\right) \\ && \end{eqnarray*}
\begin{eqnarray*}&&\\ 5x &=&\pi -\sin ^{-1}\left( \displaystyle \frac{1}{7}\rig... ...& \\ x &\approx &0.599649016937\ \mbox{ radians } \\ && \\ && \end{eqnarray*}

The exact solutions are $x=\displaystyle \frac{1}{5}\sin ^{-1}\left( \displaystyle \frac{1}{7}\right) $and $x=\displaystyle \frac{\pi }{5}-\displaystyle \frac{1}{5}\sin ^{-1}\left( \displaystyle \frac{1}{7}\right) .\bigskip\bigskip\bigskip $

The period of the sin $\left( 5x\right) $ function is $\displaystyle \frac{2\pi }{5}.$This means that the values will repeat every $\displaystyle \frac{2\pi }{5}$ radians in both directions. Therefore, the exact solutions are $x=\displaystyle \frac{1}{5}\sin ^{-1}\left( \displaystyle \frac{1}{7}\right) \pm n\left( \displaystyle \frac{2\pi }{5}\right) $ and $x= \displaystyle \frac{\pi }{5}-\displaystyle \frac{1}{5}\sin ^{-1}\left( \displaystyle \frac{1}{7}\right) \pm n\left( \displaystyle \frac{2\pi }{5}\right) $ where n is an integer.


The approximate solutions are $x\approx 0.0286695137811\pm n\left( \displaystyle \frac{ 2\pi }{5}\right) $ and $x\approx 0.599649016937\pm n\left( \displaystyle \frac{2\pi }{5} \right) $ where n is an integer.



These solutions may or may not be the answers to the original problem. You much check them, either numerically or graphically, with the original equation.



Numerical Check:


Check the answer x=0.028669513781


Since the left side equals the right side when you substitute 0.028669513781 for x, then 0.028669513781 is a solution.




Check the answer x=0.599649016937


Since the left side equals the right side when you substitute $x\approx 0.599649016937$ for x, then $x\approx 0.599649016937$ is a solution.



Graphical Check:


Graph the equation

$f(x)=7\sin \left( 5x\right) -1.$

Note that the graph crosses the x-axis many times indicating many solutions.


Note that it crosses at 0.028669513781. Since the period is $\displaystyle \frac{2\pi }{ 5}\approx 1.256637$, it crosses again at 0.028669513781+1.256637=1.285306and at 0.028669513781+2(1.256637)=2.54194, etc.


The graph crosses at 0.599649016937. Since the period is $\displaystyle \frac{2\pi }{ 5}\approx 1.256637$, it will cross again at $0.599649016937+\left( 1.256637\right) =1.856286$ and at 0.599649016937+2(1.256637)=3.112923, etc $ .\bigskip\bigskip $


If you would like to work another example, click on Example.


If you would like to test yourself by working some problems similar to this example, click on Problem.


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